Tag: maths

Let $F,H$ and $C$ denote the no. of children who play Football, Hockey and Cricket respectively.
Given $n(F)=35,n(H)=22,n(C)=25$ and $n(F\cap C\cap H^{\prime})=7,n(F\cap H\cap C^{\prime})=3,n(F\cap C)=12$
but $n(F\cap C\cap H^{\prime})=n(F\cap C)-n(F\cap C\cap H)$
$\Rightarrow 7=12-n(F\cap C\cap H)$
$\Rightarrow n(F\cap C\cap H)=5$
$\therefore$ No of children who play all three games is $5$
Hence, option A

$A _1={\pi}{r _1}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s-a)^{2}}$

Let $x-53651=9999$
Then, $x=9999+53651=63650$

The value of $+7(-2)+(-8)+(+3) $

Product of x and a = xa

$x+3699+1985-2047=31111$
$\Rightarrow$ $x+3699+1985=31111+2047$
$\Rightarrow$ $x+5684=33158$
$\Rightarrow$ $x=33158-5684=27474$

Let $4300731-x=2535618$
Then, $x=4300731-2535618=1765113$

1) The first thing we do is write these numbers one on top of the other.

347657
238294

2) Now we just start subtracting vertically, and whenever our number is negative we represent it as a vinculum number.

347657
238294

3) This next subtraction is 7 – 8 = -1…so we just write this as 1 and continue on.

347657
238294
1 1 1 443

4) The ‘1‘ and ‘4‘ here are considered as two separate groups (since theres a non-vinculum number in-between them), so each is subtracted from 10. This has the effect of reducing the “previous one” by one.

109363